- Show that
\[ \int_ {0} ^ {1} \log (\sin \pi x) d x = - \log 2. \]
[Hint: Use the contour shown in Figure 9.]
Figure 9. Contour in Exercise 9
Solution
The endpoint singularities are integrable because \[ \log(\sin t)\sim\log t\quad(t\downarrow0), \qquad \log(\sin t)\sim\log(\pi-t)\quad(t\uparrow\pi). \] Let \[ I=\int_0^{\pi/2}\log(\sin t)\,dt. \] By \(t\mapsto\pi/2-t\), \[ I=\int_0^{\pi/2}\log(\cos t)\,dt. \] Therefore \[ 2I=\int_0^{\pi/2}\log(\sin t\cos t)\,dt =\int_0^{\pi/2}\log\left(\frac12\sin 2t\right)dt. \] Thus \[ 2I=-\frac\pi2\log2+\frac12\int_0^\pi\log(\sin u)\,du. \] By symmetry about \(\pi/2\), the last integral equals \(2I\). Hence \[ 2I=-\frac\pi2\log2+I, \qquad I=-\frac\pi2\log2. \] Finally, with \(t=\pi x\), \[ \int_0^1\log(\sin\pi x)\,dx =\frac1\pi\int_0^\pi\log(\sin t)\,dt =\frac{2I}{\pi} =-\log2. \]